Answer to Question #104103 in Calculus for Akshay Kumar

Question

Find an expression for the function whose graph consists of the line segment from
the point )2,2 (− to the point )0,1 (− together with the top half of the circle with
centre at the origin and radius 1.

Answer 1

Find an expression for the function whose graph consists of the line segment from the point (-2, 2) to the point (-1, 0) together with the top half of the circle with center the origin and radius 1

Solution:

To determine:

An expression function for the graph which satisfies the given condition:

Given:

The graph has a line segment connecting (-2,2) and (-1,0) and it consists of a top half of the circle with center (0,0) and radius 1.

Calculation:

Find the slop of the line segment joining the points (-2,2) and (-1,0) as follows:

$m={y_2-y_1 \above{2pt} x_2-x_1}$

$m={0-2 \above{2pt} -1-(-2)}$

$m={-2 \above{2pt} 1}$

$m=-2$

Thus, the slope of the line segment is $m=-2$ .

Find the y-intercept of the line segment joining the points (-2,2) and (-1,0) as follows

$y=mx+c$

$0=(-2)(-1)+c$ m=-2

$0=2+c$

$c=-2$

Thus , y intercept is c=-2.

The equation of the line segment is $y=-2x-2$ for $-2\leqslant x<-1$ .

Find the equation of the circle:

The equation of the circle of radius 1 centered on the origin is $x^2+y^2=1$

We can then solve for y

$x^2+y^2=1 \iff$ $y^2=1-x^2\iff$ $y=\sqrt{1-x^2}$

To only include the top half of the circle we only take the positive root .

The equation of the circle is $\sqrt{x^2-1}$ for -1$\leqslant x\leqslant1$

$f(x)=\begin{cases} -2x-2 &\text{ on} [-2,1) \\ \sqrt{1-x^2} &\text{on} [-1,1] \end{cases}$

Here is a graph;